Question

For the purposes of this question, we will make up a name for a particular probability distribution. Say the random variable X has the July distribution with parameter b, if its moment generating function (MGF) is: My(t) = (5t – 1)-26 where b is a positive integer. For short we can write, ~ July(b). (a) If X ~ July(b), then use the MGF above to find E[X]. (b) If X ~ July(b) and Y ~ July(c) are independent, then what is the MGF of X+Y? What distribution does X + Y have?

Answer 1

is (5t-11-26 Given that the random variable_X_has_the_July dis!" with parameter bo if it's MGE My(t) = where, b is any positive integer. ..Xn July (b'). want to compute E[x] E[x] ly's @ we a mx (t) =0 at g ot (56-1)-26 tto tto -26 (57-1)-26-7 (5) 10 b (510)-1)-25 -10 b (-1)-26-1 Elx] -lob 2 btl E[x]=_-106 L-1² 2bti) 6 we Х xo Tely (b) 5 ya Julyec) both independent. want to computę MGE O and distribution of (x+y). since July (b). mx (t)=(5-1) -25 In July co my ct (Śt-1) X & Y are independent Mxty (t) = Mx17), My I (st) - 26. 26 (5t-1)-26 (5t-10-25-2

Mxty (t) est-u 2b-20 (5E-1 2 torcy myty (4 Mixtylt) (54-1-2747) we know that if two rv. is same then the distribution of that that both ov. ov. is same .. (x+y) July (btc) - Answersa ca E[x]- -10b 26+) டு nghof (x+y) is Mxty (tha (57-1) (t). Istu - 24th Distribution of exty) is parameter (b+c), July with jie X+Y N July (bte)